If a set $S$ has a proper subset $A$ such that there is a bijection from $A$ to $S$ we know that $S$ cannot be finite. Now, if a vector space $V$ has a proper subspace $W$ isomorphic to itself, then the dimension of $V$ need to be infinite?
I think the answer is yes and I've supplied the following proof: let $\alpha \subset W$ be a basis of the subspace. Since there's an isomorphism $\phi : W \to V$ and since isomorphisms transforms basis into basis, we know that $\phi(\alpha)\subset V$ is a basis of $V$ and that $\phi | \alpha : \alpha \to \phi(\alpha)$ is a bijection between those bases as sets. Now, we know from the proof that every vector space has a basis that we can produce a basis for $V$ from the set $\alpha$. In that case, let $\beta$ be that basis. We know that any two basis of $V$ has the same cardinality and we know that $\alpha \subset \beta$ and that $|\alpha| = |\phi(\alpha)|$ (because $\phi$ is a bijection between those) and we know that $|\phi(\alpha)| = |\beta|$ because those two are basis of $V$. Hence, the set $\beta$ has the same cardinality as it's proper subset $\alpha$ (which we know to be proper because $W$ is a proper subspace). In that case, since these are just sets, we must conclude that $\beta$ is infinite.
Now, I'm a little unsure of this proof. Is this proof correct? Is the proposition true?
Thanks very much in advance!
Best Answer
Suppose $\dim W<\infty$. Since $|\alpha| = |\phi(\alpha)|$ and $\phi(\alpha)$ is linearly independent, $\phi(\alpha)$ is a basis of $W$ (this is the property of finite dimensional spaces). Hence $V=W$.